"Unpredictable Solution" Problem (Part 2)

$\begin{cases} xy-2x-y = 10 \space ...(1) \\ yz-3y-2z = 14 \space ...(2) \\ zx-z-3x=12 \space ...(3) \end{cases}$

$(1) + 2$ gives $xy -2x -y + 2 = 12 \space \Rightarrow \space (x-1)(y-2) = 12 \space ...(4)$ .

$(2) + 6$ gives $yz -3y -2z + 6 = 20 \space \Rightarrow \space (y-2)(z-3) = 20 \space ...(5)$ .

$(3) + 3$ gives $zx - z - 3x + 3 = 15 \space \Rightarrow \space (z-3)(x-1) = 15 \space ...(6)$ .

$(4) \times (5) \times (6)$ gives $(x-1)(y-2)(z-3) = 60 \space ...(7)$ .

$(7) : (4)$ gives $z = 8$ .

$(7) : (5)$ gives $x = 4$ .

$(7) : (6)$ gives $y = 6$ .

Then, we have $(x+y+z)^2 + x² + y² + z² = 18² + 4² + 6² + 8²$ .

Hence, our final answer is $\boxed{440 }$ .

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Otherwise, by the system of the equation above, we can find that $(-2,-2,-2)$ complete the system that is given. But, since $(x,y,z)$ must be positive, we can use this method.